Question: Evaluate the Maclaurin series. $3-\dfrac{{{3\pi }^{4}}}{2!}+\dfrac{{3{\pi }^{8}}}{4!}-...+{{3\left( -1 \right)}^{n}}\dfrac{{{\pi }^{4n}}}{\left( 2n \right)!}+...$ Choose 1 answer: Choose 1 answer: (Choice A) A $3$ (Choice B) B $-3$ (Choice C) C $3\cos ( \pi^2 )$ (Choice D) D $-3\cos ( \pi^2 )$
Answer: Note that the series has terms that alternate in sign, has even factorials in the denominator, and has even powers of $~\pi~$ in the numerator. This suggests that we work with the Maclaurin series for $~\cos x\,$. Recall that $\cos x=1-\frac{{{x}^{2}}}{2!}+\frac{{{x}^{4}}}{4!}-...+{{\left( -1 \right)}^{n}}\frac{{{x}^{2n}}}{\left( 2n \right)!}+...$ It follows that $3\cos (x^2)=3-\frac{3{{x}^{4}}}{2!}+\frac{3{{x}^{8}}}{4!}-...+{3{\left( -1 \right)}^{n}}\frac{{{x}^{4n}}}{\left( 2n \right)!}+...$ Hence, the given series converges to $~3\cos ( \pi^2 )\,$.